The club was given materials to build 160 feet of fence. If they used the auditorium wall for one side of their garden, what size rectangle should they make to have the largest area? If w represents the width of the rectangle and is perpendicular to the auditorium wall, what is the largest value possible for w? Why?
in these kinds of problems, the largest area is always achieved when the available fencing is divided equally among lengths and widths. In this case, the maximum area is found when the garden is 80x40.
How did you get 80x40?
To determine the largest possible value for the width of the rectangle (w), we need to optimize the area of the rectangle given the constraint of using 160 feet of fence material.
Let's break down the problem into steps:
Step 1: Express the perimeter in terms of w
The perimeter of the rectangle is given as 160 feet. Since one side is the auditorium wall and the other three sides will be built using the materials, we have:
Perimeter = 160 feet = w + 2L,
where L represents the length of the rectangle.
Step 2: Express L in terms of w
Since one side of the rectangle is the auditorium wall, the other side parallel to the wall will have length L as well. Therefore, we can express L in terms of w as:
L = w.
Step 3: Express the area of the rectangle in terms of w
The area (A) of the rectangle is given by the formula:
Area = w * L,
Substituting L = w, we get:
Area = w * w = w^2.
Step 4: Optimize the area
To find the largest possible value of w, we need to find the value of w that optimizes the area A = w^2.
Given the perimeter equation from Step 1, we can isolate L as:
L = (160 - w) / 2.
Now substitute this value of L into the area equation:
Area = w * L = w * [(160 - w) / 2] = (160w - w^2) / 2.
Step 5: Maximize the area function by taking derivative and solving for critical points
To maximize the area function, we differentiate it with respect to w:
d(Area) / d(w) = (160 - 2w) / 2 = 80 - w.
Setting this derivative equal to zero gives us the critical point:
80 - w = 0,
w = 80.
Step 6: Evaluate the second derivative to confirm the maximum value
Taking the second derivative of the area function:
d^2(Area) / d(w)^2 = -2,
Since the second derivative is negative, this confirms that w = 80 is a maximum.
So, the largest possible value for the width of the rectangle (w) is 80 feet.